Funky Solutions to Tricky Limit Problems?

[matineesuxxx]

funky limit solutions??

hey guys, this is my first post, and absolutely love this site! Anyways, I've been studying some calculus to get ahead of the game before university and I came across a few limit questions that I solved, but I don't really know HOW I solved them and by that, I mean, with some of the substitutions, I got some limits that seem intuitive to me, but I don't really know why they are that way.. Forgive my latex if it doesn't end up lookin all pretty, I am pretty new to it. So I was hoping anybody could take a look at these 3 questions and let me know whatsup?
1) \begin{equation*}
\lim_{x\rightarrow \infty}x\,{\sin{\frac{1}{x}}}\quad \text{let}\,a=\frac{1}{x}\\\\\quad\therefore \lim_{x\rightarrow\infty}a= \lim_{a \rightarrow 0}a=0\\\\\qquad \therefore \lim_{x\rightarrow\infty}x\sin\frac{1}{x}=\lim_{a \rightarrow 0}\frac{\sin(a)}{a}=1\\\\\quad\therefore\lim_{x \rightarrow \infty}x\sin{\frac{1}{x}}=1
\end{equation*}

2) \begin{equation*}
\lim_{x\rightarrow 1}\frac{\sin(x-1)}{x^2+x-2}\\\\=\lim_{x\rightarrow1}\frac{\sin(x-1)}{(x+2)(x-1)}\\\\\text{let}\,a=x-1\qquad \therefore\,a+3=x+2\\\\ *\lim_{x\rightarrow 1}(x-1)=0\Rightarrow \lim_{a\rightarrow0}a=0\\\\ \therefore \lim_{x\rightarrow}\frac{\sin(x-1)}{(x+2)(x-1)}=\lim_{a\rightarrow0}\frac{\sin(a)}{a(a+3)}\\\\=(\lim_{a\rightarrow 0}\frac{1}{a+3}) (\lim_{a\rightarrow 0}\frac{\sin(a)}{a})\\\\=\frac{1}{3}
\end{equation*}

\begin{equation*}
\lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta+\tan \theta}\\\\\\=\lim_{\theta \rightarrow 0}\frac{\cos\theta \sin\theta}{\theta \cos \theta+\sin \theta}\\\\\\= \lim_{\theta \rightarrow 0}\bigg( \Big( \frac{\cos\theta\sin\theta}{\theta\cos\theta+\sin\theta}\Big)^{-1} \bigg)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta\cos\theta+\sin\theta}{\cos\theta \sin\theta}\Big)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta\cos \theta}{\cos \theta \sin \theta}+\frac{\sin\theta}{\cos\theta\sin\theta} \Big)^{-1}\\\\\\=\lim_{\theta \rightarrow 0}\Big(\frac{\theta}{\sin\theta}+\frac{1}{\cos\theta}\Big)^{-1}\\\\\\ *\lim_{\theta\rightarrow0}\Big(\frac{\theta}{\sin \theta}\Big)^{-1}=\lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1 *\\\\\\= \Bigg( \lim_{\theta\rightarrow 0}\frac{\theta}{\sin\theta}+\lim_{\theta \rightarrow 0}\frac{1}{\cos\theta}\Bigg)^{-1}\\\\\\=\big(1+1)^{-1}=\frac{1}{2}
\end{equation*}

so sorry about the bad latex! I used that codecogs site, and I guess it is just a tad different?

 

[Dick]

The solutions all look good. Can you be a little more specific about what your question is?

 

[matineesuxxx]

The solutions all look good. Can you be a little more specific about what your question is?

well, in my first one, with the:
\begin{equation*}
\lim_{x \rightarrow \infty} \frac{1}{x}=0 \Rightarrow \lim_{x \rightarrow \infty}a=0 = \lim_{a \rightarrow 0} a = 0
\end{equation*}

that is what I felt like I should do, but I was concerned about whether or not it was faulty logic?, (and I felt the same concern with that method again in the second problem), also I just felt like there should be a much much simpler way of solving the last one, because I tend to complicate things. sorry for not making my concerns known earlier :)

 

[Ray Vickson]

well, in my first one, with the:
\begin{equation*}
\lim_{x \rightarrow \infty} \frac{1}{x}=0 \Rightarrow \lim_{x \rightarrow \infty}a=0 = \lim_{a \rightarrow 0} a = 0
\end{equation*}

that is what I felt like I should do, but I was concerned about whether or not it was faulty logic?, (and I felt the same concern with that method again in the second problem), also I just felt like there should be a much much simpler way of solving the last one, because I tend to complicate things. sorry for not making my concerns known earlier :)

Why would you think it could be faulty? Fact: as $x \to \infty$, $1/x \to 0$.

 

[matineesuxxx]

Why would you think it could be faulty? Fact: as $x \to \infty$, $1/x \to 0$.

Ok, well that makes me feel a lot better to get your input. I was told by a teacher that it was "unreasonable" to say that, then to go on and say,

\begin{equation*}
\lim_{x \rightarrow \infty}x\sin(\frac{1}{x})=\lim_{a \rightarrow 0}\frac{\sin(a)}{a}
\end{equation*}

I'm new to calculus, so I'm not very confident yet. So thanks a bunch!

 

[Ray Vickson]

Ok, well that makes me feel a lot better to get your input. I was told by a teacher that it was "unreasonable" to say that, then to go on and say,

\begin{equation*}
\lim_{x \rightarrow \infty}x\sin(\frac{1}{x})=\lim_{a \rightarrow 0}\frac{\sin(a)}{a}
\end{equation*}

I'm new to calculus, so I'm not very confident yet. So thanks a bunch!

I think your teacher's statement is misleading: your argument is a good one, although maybe you need to state explicitly that you are replacing $1/x$ by $a.$